$A$ triangular shaped wire carrying $10 A$ current is placed in a uniform magnetic field of $0.5 T$,as shown in the figure. The magnetic force on segment $CD$ is $.... N$ (Given $BC = CD = BD = 5 cm$).

The resultant force on the current loop $PQRS$ due to a long current-carrying conductor is:

The figure shows a conducting loop $ABCDA$ placed in a uniform magnetic field perpendicular to its plane. The part $ABC$ is the $(3/4)^{th}$ portion of the square of side length $l$. The part $ADC$ is a circular arc of radius $R$. The points $A$ and $C$ are connected to a battery which supplies a current $I$ to the circuit. The magnetic force on the loop due to the field $B$ is

$A$ circular current loop of radius $a$ is placed in a radial magnetic field $B$ as shown in the figure. The net force acting on the loop is

$A$ current loop $ABCD$ is held fixed on the plane of the paper as shown in the figure. The arcs $BC$ (radius $= b$) and $DA$ (radius $= a$) of the loop are joined by two straight wires $AB$ and $CD$. $A$ steady current $I$ is flowing in the loop. The angle made by $AB$ and $CD$ at the origin $O$ is $30^o$. Another straight thin wire with a steady current $I_1$ flowing out of the plane of the paper is kept at the origin. Due to the presence of the current $I_1$ at the origin:

$A$ square current-carrying loop is suspended in a uniform magnetic field acting in the plane of the loop. If the force on one arm of the loop is $\overrightarrow{F}$,the net force on the remaining three arms of the loop is